Select the pair, which are dimensionally alike among the following

Ⅰ. Product of force and time

Ⅱ. Product of momentum and time

Ⅲ. Product of Ariel velocity and linear density

Ⅳ. Product of work and time

Understanding Dimensional Analysis

Dimensional analysis is a powerful tool in physics used to check the consistency of equations and to derive relationships between physical quantities. Every physical quantity can be expressed in terms of fundamental dimensions such as Mass (M), Length (L), and Time (T).

To determine if a pair of quantities are dimensionally alike, we first need to find the dimensions of each individual quantity involved in their product. We will then compare the resulting dimensions.

Fundamental Dimensions and Key Quantities

Here are the fundamental dimensions and the dimensions of common physical quantities relevant to this problem:Calculating Dimensions for Each Product

Let's calculate the dimensions for each of the four given products:

I. Product of force and time

- Force (F): The dimension of force is ${\text{[M][L][T]}}^{-2}$.
- Time (t): The dimension of time is $\text{[T]}$.
- Product Dimension: $\text{F}\times \text{t}=({\text{[M][L][T]}}^{-2})\times (\text{[T]})={\text{[M][L][T]}}^{-2+1}={\text{[M][L][T]}}^{-1}$.
- This dimension is also known as the dimension of Impulse or Momentum.

II. Product of momentum and time

- Momentum (p): The dimension of momentum is ${\text{[M][L][T]}}^{-1}$.
- Time (t): The dimension of time is $\text{[T]}$.
- Product Dimension: $\text{p}\times \text{t}=({\text{[M][L][T]}}^{-1})\times (\text{[T]})={\text{[M][L][T]}}^{-1+1}={\text{[M][L][T]}}^0=\text{[M][L]}$.

III. Product of aerial velocity and linear density

The term "aerial velocity" in this context is likely a typographical error for "areal velocity". Areal velocity refers to the rate at which area is swept out by a line segment connecting a reference point to a moving body. Its dimension is Area divided by Time.

- **Areal Velocity ($\frac{\text{dA}}{\text{dt}}$):** The dimension of area is ${\text{[L]}}^2$, and time is $\text{[T]}$. So, the dimension of areal velocity is ${\text{[L]}}^2{\text{[T]}}^{-1}$.
- **Linear Density ($\lambda$):** The dimension of linear density (mass per unit length) is ${\text{[M][L]}}^{-1}$.
- Product Dimension: $\text{Areal Velocity}\times \text{Linear Density}=({\text{[L]}}^2{\text{[T]}}^{-1})\times ({\text{[M][L]}}^{-1})={\text{[M][L]}}^{2-1}{\text{[T]}}^{-1}={\text{[M][L][T]}}^{-1}$.

If "aerial velocity" were interpreted simply as "velocity" (${\text{[L][T]}}^{-1}$), the product dimension would be ${\text{[L][T]}}^{-1}\times {\text{[M][L]}}^{-1}={\text{[M][T]}}^{-1}$, which would not match any other option. Therefore, "areal velocity" is the intended meaning to find a match.

IV. Product of work and time

- Work (W): The dimension of work is ${\text{[M][L]}}^2{\text{[T]}}^{-2}$.
- Time (t): The dimension of time is $\text{[T]}$.
- Product Dimension: $\text{W}\times \text{t}=({\text{[M][L]}}^2{\text{[T]}}^{-2})\times (\text{[T]})={\text{[M][L]}}^2{\text{[T]}}^{-2+1}={\text{[M][L]}}^2{\text{[T]}}^{-1}$.

Comparison of Dimensions

Let's summarize the dimensions we have calculated for each product:Comparing the calculated dimensions:

- Product I has the dimension ${\text{[M][L][T]}}^{-1}$.
- Product II has the dimension $\text{[M][L]}$.
- Product III has the dimension ${\text{[M][L][T]}}^{-1}$.
- Product IV has the dimension ${\text{[M][L]}}^2{\text{[T]}}^{-1}$.

From the comparison, it is clear that Product I and Product III have the same dimensions (${\text{[M][L][T]}}^{-1}$). Therefore, they are dimensionally alike.